hdu 5742 It's All In The Mind（2016多校第二场）

It's All In The Mind

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 505    Accepted Submission(s):
225

Problem Description

Professor Zhang has a number sequence a1,a2,...,an . However, the sequence is not complete and some elements are missing.
Fortunately, Professor Zhang remembers some properties of the
sequence:

1. For every i∈{1,2,...,n} , 0≤ai≤100 .
2. The sequence is non-increasing, i.e. a1≥a2≥...≥an .
3. The sum of all elements in the sequence is not zero.

Professor
Zhang wants to know the maximum value of a1+a2∑ni=1ai among all the possible sequences.

 

Input

There are multiple test cases. The first line of input
contains an integer T , indicating the number of test cases. For each test case:

The first
contains two integers n and m (2≤n≤100,0≤m≤n) -- the length of the sequence and the number of known elements.

In the
next m lines, each contains two integers xi and yi (1≤xi≤n,0≤yi≤100,xi<xi+1,yi≥yi+1) , indicating that axi=yi .

 

Output

For each test case, output the answer as an irreducible
fraction "p /q ", where p , q are integers, q>0 .

 

Sample Input

2

2 0

3 1

3 1

 

Sample Output

1/1

200/201

 

Author

zimpha

 

题意：求a1+a2/a1+a2+..+an的最大值，其中只给出部分的值，ax=y;

 

水题，保证a1,a2最大，后面的尽量小就好，需要注意的是，这是非递增数列，后面有数赋值，会影响前面的数的取值。

 

附上代码：

 

 #include <iostream>
 #include <cstdio>
 using namespace std;
 int a[];
 int _gcd(int x,int y) ///求最大公约数
 {
     int z;
     if(x<y) z=x,x=y,y=z;
     while(y)
     {
         z=x%y;
         x=y;
         y=z;
     }
     return x;
 }
 int main()
 {
     int T,i,j,n,m;
     scanf("%d",&T);
     while(T--)
     {
         int x,y,t=;
         scanf("%d%d",&n,&m);
         for(i=; i<=n; i++)  ///初始化全为-1
             a[i]=-;
         for(i=; i<m; i++)
         {
             scanf("%d%d",&x,&y);
             a[x]=y;
         }
         if(n==||m==)   ///若只有两个数，或者不给任何数赋值，都能输出最大的1/1
         {
             printf("1/1\n");
             continue;
         }
         int nn=,mm=,fail=;
         for(i=n; i>=; i--)  ///因为前面的小于等于后面的，因此从后面向前面循环
         {
                 if(a[i]!=-) ///如果a[i]被赋值，记录此时的a[i]，并标记已出现赋值的数
                 {
                     t=a[i];
                     fail=;
                     mm+=t;
                 }
                 else
                 {
                     if(!fail)  ///若没有出现已赋值的数，可以定义为后面的数始终为0
                     {
                         mm+=;
                     }
                     else   ///若已出现，则只能小于等于这个数
                     {
                         mm+=t;
                     }
                 }
         }
         if(a[]==-) ///a1和a2期望越大越好，若都无赋值，就都取100，若有赋值，则去赋值的数，且a2小于a1
         {
             nn+=,mm+=;
             if(a[]==-)
                 nn+=,mm+=;
             else
                 nn+=a[],mm+=a[];
         }
         else
         {
             nn+=a[],mm+=a[];
             if(a[]==-)
                 nn+=a[],mm+=a[];
             else
                 nn+=a[],mm+=a[];
         }
         int dd=_gcd(nn,mm);
         printf("%d/%d\n",nn/dd,mm/dd);
     }
     return ;
 }